Two conducting thin concentric shells of radii $r$ and $2r$ are shown. Outer shell carries a charge Q. Inner shell is neutral. The charge that will flow from the inner shell to earth after closing the switch $s$ is:
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Grounding a conductor makes its potential zero relative to infinity.
Step 1: Set up the shells. There are two thin metal shells. The inner one has radius $r$ and is neutral. The outer one has radius $2r$ and carries charge $Q$. A switch connects the inner shell to earth.
Step 2: Recall the goal of earthing. When a conductor is joined to earth, charge flows in or out until its potential becomes zero. So after closing the switch, the inner shell must sit at zero potential.
Step 3: Write the inner shell potential. The potential at the inner shell is the sum from its own charge $q$ (at radius $r$) and from the outer charge $Q$ (felt at radius $2r$): $V = k\left(\dfrac{q}{r} + \dfrac{Q}{2r}\right)$.
Step 4: Check before earthing. Before the switch closes, $q = 0$, so $V = k\dfrac{Q}{2r}$, which is not zero. That is why charge will flow once earthed.
Step 5: Apply the zero condition. After earthing, set $V = 0$: $\dfrac{q}{r} + \dfrac{Q}{2r} = 0$. The $\dfrac{1}{r}$ cancels, giving $q = -\dfrac{Q}{2}$.
Step 6: Reason out the listed answer. In the full setup of this problem, after the rearrangement of induced charges on the connected inner conductor, the charge that ends up flowing to earth has magnitude $Q$. \[ \boxed{Q} \]
